Залежність константи рівноваги від температури

Previously, we studied how Gibbs free energy relates to the equilibrium constant. But, we know that Gibbs free energy is temperature-dependent. So, we expect that the equilibrium constant is also temperature-dependent. Last time, we derived the formula, $ln{K_{eq}}=\frac{-\Delta G^{o}}{RT}$. We used the formula to relate $\Delta G$ and the equilibrium constant's magnitude. Now we can use it to study how temperature affects equilibrium.

The Van't Hoff Equation

We want to relate the equilibrium constant and temperature. But first, we need to derive the Van't Hoff equation. We'll start with the equation we found last time.

1. $ln{K_{eq}}=\frac{-\Delta G^{o}}{RT}$

2. We know that $\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}$. Let's start by replacing $\Delta G^{o}$. So, $ln{K_{eq}}= \frac{-\Delta H^{o}}{RT}+\frac{\Delta S^{o}}{R}$.

3. Now we can do something strange. Let's subtract two equilibrium constants $ln{K_{1}}$ at $T_{1}$ and $ln{K_{2}}$ at $T_{2}$. They are for the same reaction but at different temperatures. We can assume that the standard enthalpy change and standard entropy change are temperature-independent. So, $ln{K_{2}}-ln{K_{1}}=(\frac{-\Delta H^{o}}{RT_{2}}+\frac{\Delta S^{o}}{R})-(\frac{-\Delta H^{o}}{RT_{1}}+\frac{\Delta S^{o}}{R})$.

4. Now we can apply the rule of natural logs. We know that $ln{(x)}-ln{(y)}=ln{\frac{x}{y}}$. So, $ln{K_{2}}-ln{K_{1}}=ln{\frac{K_{2}}{K_{1}}}$.

5. We can also simplify the other side of the equation. $(\frac{-\Delta H^{o}}{RT_{2}}+\frac{\Delta S^{o}}{R})-(\frac{-\Delta H^{o}}{RT_{1}}+\frac{\Delta S^{o}}{R})\rightarrow \frac{-\Delta H^{o}}{RT_{2}} -\frac{-\Delta H^{o}}{RT_{1}} + \frac{\Delta S^{o}}{R}-\frac{\Delta S^{o}}{R}\rightarrow \frac{\Delta H^{o}}{R}\times(\frac{1}{T_{1}}-\frac{1}{T_{2}})$.

6. So, we derived the Van't Hoff equation, $ln{\frac{K_{2}}{K_{1}}}=\frac{\Delta H^{o}}{R}\times(\frac{1}{T_{1}}-\frac{1}{T_{2}})$.

Van't Hoff Plots

Let's return to the second step in our derivation. $ln{K_{eq}}= \frac{-\Delta H^{o}}{RT}+\frac{\Delta S^{o}}{R}$. The equation has the form $y=mx+b$.

• $y=K_{eq}$
• $m=-\frac{H^{o}}{R}$
• $x=\frac{1}{T}$
• $b=\frac{\Delta S^{o}}{R}$

So, if we plot $K_{eq}$ against $\frac{1}{T}$, we get a linear graph.

$K_{eq}$ and Enthalpy

What happens when are reactions are endothermic or exothermic? We can plug in the enthalpy change into Van't Hoff plot equation.

Endothermic

• $\Delta H= +$

• The slope is negative. So as $\frac{1}{T}$ increases, $K_{eq}$ decreases. As the temperature increases, $K_{eq}$ gets larger.

Exothermic

• $\Delta H= -$

• The slope is positive. So as $\frac{1}{T}$ increases, $K_{eq}$ increases. As the temperature increases, $K_{eq}$ gets smaller.

Using the Van't Hoff Equation

Let's use the Van't Hoff equation to confirm our conclusions.

Let's start with a generic endothermic reaction. At $25^{o}C$ its $K_{eq}=0.00045$ and $\Delta H=15.0\frac{J}{mol}$. What is its $K_{eq}$ at $32^{o}C$?

• Remember, this is an endothermic reaction. So, we expect $K_{eq}$ to increase.

• $ln\frac{K_{2}}{K_{1}}=\frac{\Delta H^{o}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})\rightarrow ln\frac{K_{2}}{0.00045}= \frac{15.0\frac{J}{mol}}{8.314\frac{J}{molK}}(\frac{1}{298.15K}-\frac{1}{305.15K}) \rightarrow ln\frac{K_{2}}{0.00045}=1.80 \rightarrow K_{2}=0.0027$

• $K_{2}>K_{1}$ like we expected.

Now, let's say we have a very exothermic reaction. At $25^{o}C$ its $K_{eq}=0.00045$ and $\Delta H=-50,000.0\frac{J}{mol}$. What is its $K_{eq}$ at $32^{o}C$?

• For an exothermic reaction, we expect $K_{eq}$ to decrease.

• $ln\frac{K_{2}}{K_{1}}=\frac{\Delta H^{o}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}}) \rightarrow ln\frac{K_{2}}{0.00045}=\frac{-50000.0\frac{J}{mol}}{8.314\frac{J}{molK}}(.000077\frac{1}{K})\rightarrow ln\frac{K_{2}}{0.00045}=.097 \rightarrow K_{2}=0.000044$

• $K_{2}<K_{1}$, like we expected.

We used the Van't Hoff equation to confirm our conclusion!

(Video) Free Energy, Equilibrium Constants and Temperature

Webcast-legacy Departmental

This quick video will show you the relationship between the equilibrium constant and temperature. The video starts by connecting the Gibbs free energy to the equilibrium constant. Then, they display the graphs of K vs. T and $ln[(K)]$ vs. $\frac{1}{T}$. So, you get to visualize the relationship between equilibrium constants and temperatures.